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Commit 28998bf

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feat: add solutions to lc problem: No.1304
1 parent 98852a2 commit 28998bf

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6 files changed

+103
-6
lines changed

6 files changed

+103
-6
lines changed

‎solution/1300-1399/1304.Find N Unique Integers Sum up to Zero/README.md

Lines changed: 34 additions & 1 deletion
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@@ -127,7 +127,7 @@ func sumZero(n int) []int {
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```ts
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function sumZero(n: number): number[] {
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const ans=new Array(n).fill(0);
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const ans:number[] = Array(n).fill(0);
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for (let i = 1, j = 0; i <= n / 2; ++i) {
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ans[j++] = i;
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ans[j++] = -i;
@@ -136,6 +136,24 @@ function sumZero(n: number): number[] {
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}
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```
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#### Rust
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```rust
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impl Solution {
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pub fn sum_zero(n: i32) -> Vec<i32> {
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let mut ans = vec![0; n as usize];
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let mut j = 0;
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for i in 1..=n / 2 {
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ans[j] = i;
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j += 1;
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ans[j] = -i;
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j += 1;
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}
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ans
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}
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}
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```
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<!-- tabs:end -->
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<!-- solution:end -->
@@ -215,6 +233,21 @@ function sumZero(n: number): number[] {
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}
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```
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#### Rust
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```rust
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impl Solution {
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pub fn sum_zero(n: i32) -> Vec<i32> {
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let mut ans = vec![0; n as usize];
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for i in 1..n {
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ans[i as usize] = i;
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}
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ans[0] = -(n * (n - 1) / 2);
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ans
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}
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}
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```
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<!-- tabs:end -->
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<!-- solution:end -->

‎solution/1300-1399/1304.Find N Unique Integers Sum up to Zero/README_EN.md

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@@ -57,7 +57,11 @@ tags:
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<!-- solution:start -->
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### Solution 1
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### Solution 1: Construction
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We can start from 1ドル$ and alternately add positive and negative numbers to the result array. We repeat this process $\frac{n}{2}$ times. If $n$ is odd, we add 0ドル$ to the result array at the end.
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The time complexity is $O(n),ドル where $n$ is the given integer. Ignoring the space used for the answer, the space complexity is $O(1)$.
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<!-- tabs:start -->
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@@ -124,7 +128,7 @@ func sumZero(n int) []int {
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```ts
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function sumZero(n: number): number[] {
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const ans=new Array(n).fill(0);
131+
const ans:number[] = Array(n).fill(0);
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for (let i = 1, j = 0; i <= n / 2; ++i) {
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ans[j++] = i;
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ans[j++] = -i;
@@ -133,13 +137,35 @@ function sumZero(n: number): number[] {
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}
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```
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#### Rust
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```rust
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impl Solution {
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pub fn sum_zero(n: i32) -> Vec<i32> {
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let mut ans = vec![0; n as usize];
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let mut j = 0;
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for i in 1..=n / 2 {
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ans[j] = i;
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j += 1;
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ans[j] = -i;
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j += 1;
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}
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ans
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}
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}
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```
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<!-- tabs:end -->
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<!-- solution:end -->
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<!-- solution:start -->
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### Solution 2
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### Solution 2: Construction + Mathematics
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We can also add all integers from 1ドル$ to $n-1$ to the result array, and finally add the opposite of the sum of the first $n-1$ integers, which is $-\frac{n(n-1)}{2},ドル to the result array.
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The time complexity is $O(n),ドル where $n$ is the given integer. Ignoring the space used for the answer, the space complexity is $O(1)$.
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<!-- tabs:start -->
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}
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```
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#### Rust
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```rust
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impl Solution {
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pub fn sum_zero(n: i32) -> Vec<i32> {
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let mut ans = vec![0; n as usize];
243+
for i in 1..n {
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ans[i as usize] = i;
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}
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ans[0] = -(n * (n - 1) / 2);
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ans
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}
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}
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```
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<!-- tabs:end -->
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<!-- solution:end -->
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@@ -0,0 +1,13 @@
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impl Solution {
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pub fn sum_zero(n: i32) -> Vec<i32> {
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let mut ans = vec![0; n as usize];
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let mut j = 0;
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for i in 1..=n / 2 {
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ans[j] = i;
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j += 1;
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ans[j] = -i;
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j += 1;
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}
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ans
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}
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}

‎solution/1300-1399/1304.Find N Unique Integers Sum up to Zero/Solution.ts

Lines changed: 1 addition & 1 deletion
Original file line numberDiff line numberDiff line change
@@ -1,5 +1,5 @@
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function sumZero(n: number): number[] {
2-
const ans=new Array(n).fill(0);
2+
const ans: number[]= Array(n).fill(0);
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for (let i = 1, j = 0; i <= n / 2; ++i) {
44
ans[j++] = i;
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ans[j++] = -i;

‎solution/1300-1399/1304.Find N Unique Integers Sum up to Zero/Solution2.java

Lines changed: 1 addition & 1 deletion
Original file line numberDiff line numberDiff line change
@@ -7,4 +7,4 @@ public int[] sumZero(int n) {
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ans[0] = -(n * (n - 1) / 2);
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return ans;
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}
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}
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}
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Original file line numberDiff line numberDiff line change
@@ -0,0 +1,10 @@
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impl Solution {
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pub fn sum_zero(n: i32) -> Vec<i32> {
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let mut ans = vec![0; n as usize];
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for i in 1..n {
5+
ans[i as usize] = i;
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}
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ans[0] = -(n * (n - 1) / 2);
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ans
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}
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}

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